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WLOG, let the side length of one of the smaller equilateral triangles by $1$. So the overall area of one of the smaller equilateral triangles is $\frac{\sqrt{3}}{4}.$ To find the area of the shaded region (within the smaller triangle), I toyed with the idea of coord-bashing (I know, a bad method, but I didn't know what else to do), but quickly dismissed it. It would take way too much time.

I am at a loss for what to do, can someone please provide me with a hint (not solution) that will help me on my way?

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  • $\begingroup$ Hint: consider the unshaded part of the dotted triangle. The horizontal "top" side proportion is easily read off, the slanted side is not too bad, by consider the parallel edges in the diagram. $\endgroup$ – user10354138 Mar 6 at 9:26
  • $\begingroup$ For what it's worth, I would have invested about 15 minutes looking for elegant insight. That would have failed. Then I would have invested about 60 minutes (or less) in coordinate bashing. That would have succeeded. As a rule of thumb, when there is an overlooked elegant approach, the inelegant approach often simplifies more easily than one would expect. $\endgroup$ – user2661923 Mar 6 at 10:25
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EDIT: replacing hint with a solution after consent from OP

WLOG, sides of each small equilateral triangle is $1$.

As $UV$ and $XY$ are parallel to $BC$, $AM$ divides all of them in the same ratio. From that, we get

$XT = 4 \cdot \frac{3}{5} = \frac{12}{5} \implies ST = \displaystyle \frac{2}{5}$. Similarly we find $PQ = \displaystyle \frac{1}{5}$.

We also observe that $\angle QPR = \angle RST = 60^0$. So, $\triangle PRQ \sim \triangle SRT$.

Altitude from $R$ to $OP$ is $1/3$ times altitude from $S$ to $OP$.

So, Area $\triangle PQR = \displaystyle \frac{1}{15} \cdot$ Area $\triangle OPS$

$\therefore \ $ shaded part is $\frac{14}{15}$ times the dotted area.

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    $\begingroup$ @cosmo5 thank you. The problem is that the OP said they want only a hint and not a solution. I was not sure how much was a hint vs. a solution :) Yes that is the answer I get too. $\endgroup$ – Math Lover Mar 6 at 14:17
  • $\begingroup$ Ok, here is one solution I can think of now: We can find ST and PQ (by using the ratios of UV and XY you said). From there, we can find PR, RS, and QR, RT easily. From there, Heron's will give me the area of PQR. $1-\frac{[PQR]}{\sqrt{3}/4}$ is the answer. Did you guys find the same solution, or follow a different track? $\endgroup$ – Mike Smith Mar 6 at 19:51
  • $\begingroup$ @MikeSmith as we have to find ratio, it is unnecessary to use Heron's formula and find area. Use ratio of sides to say what the ratio of area will be. $\endgroup$ – Math Lover Mar 6 at 20:03
  • $\begingroup$ Please let me know if I can now add a bit of details to my answer so it will be easier for someone else to follow. $\endgroup$ – Math Lover Mar 6 at 20:22
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    $\begingroup$ Is the answer $\frac{14}{15}$? And yes, you can add some details. Thank you so much! :) $\endgroup$ – Mike Smith Mar 6 at 20:41
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As a possible helpful path, consider the (subdivided) rectangle shown, which is exactly half shaded overall and the subdivisions are $\frac 18, \frac 38, \frac 58$ and $\frac 78$ shaded.

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  • $\begingroup$ +1 for the idea. Will use elsewhere :) $\endgroup$ – cosmo5 Mar 7 at 6:28

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